Question

AC is quarter of a circle, BC is a semicircle, and ABCD is a square. Find the area of the shaded region. (Take π = 22/7)

Answer 1

Answer:

119 cm²

Step-by-step explanation:

Area of the whole circle one fourth of which is inscribed in the square is:

$=\pi r^2\\\\=(\frac{22}{7})(14)(14)\\\\= 616 cm^2$

Now one fourth of this circle causes the white part in the square ABCD

Ar(ABCD) - $\frac{1}{4} (Ar.Circle)$ = Ar(curve in ABCD)

Ar(ABCD) = 14 x 14

=196

$\frac{1}{4}Ar.Circle =( \frac{1}{4} )(616)\\= 154 cm^2$

So,  Ar(curve in ABCD) = 196 - 154

=42 cm²

Now area of the semicircle with diameter BC

BC = 14

r = 7 cm

Area of semicircle

$\frac{1}{2}( \pi r^2)\\\\= \frac{1}{2}(\frac{22}{7} )(7)(7)\\\\= 77 cm^2$

Total shaded area = Ar(curve in ABCD) + Area of semicircle

= 42 + 77

= 119 cm²